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Lecture XIII 37

Lecture XIII: Double Well Potential: Tunneling and Instantons

How can phenomena of QM tunneling be described by Feynman path integral?No semi-classical expansion!

E.g QM transition probability of particle in double well: G(a,a; t) a|eiHt/| a

Potential

x

Invertedpotential

V(x)

Feynman Path Integral:

G(a,a; t) =

q(t)=aq(0)=a

Dqexp

i

t0

dtm

2q2 V(q)

Stationary phase analysis: classical e.o.m. mq= qV only singular (high energy) solutions Switch to alternative formulation...

Imaginary (Euclidean) time Path Integral: Wick rotation t = i

N.B. (relative) sign change! V V

G(a,a; ) =

q()=aq(0)=a

Dqexp

1

0

dm

2q2 + V(q)

Saddle-point analysis: classical e.o.m. mq= +V(q) in inverted potential!solutions depend on b.c.

(1) G(a, a; )

qcl() = a(2) G(a,a; ) qcl() = a(3) G(a,a; ) qcl : rolls from a to a

Combined with small fluctuations, (1) and (2) recover propagator for single well

Invertedpotential

-a a

-V(x)

x

a

-a

x

1/

(3) accounts for QM tunneling and is known as an instanton (or kink)

Lecture Notes October 2005

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Lecture XIII 38

Instanton: classically forbidden trajectory connecting two degenerate minima

i.e. topological, and therefore particle-like

For large, qcl 0 (evident), i.e. first integral mqcl2/2 V(qcl) =

0

precise value of fixed by b.c. (i.e. )Saddle-point action (cf. WKB

dqp(q))

Sinst. =

0

dm

2q2cl + V(qcl)

0

dmq2cl =

aa

dqclmqcl =

aa

dqcl(2mV(qcl))1/2

Structure of instanton: For q a, V(q) = 12m2(q a)2 + , i.e. qcl

(qcl a)

qcl()

= a e, i.e. temporal extension set by 1

Imples existence of approximate saddle-point solutionsinvolving many instantons (and anti-instantons): instanton gas

1

a

-a

x

5

4

3

2

Accounting for fluctuations around n-instanton configuration

G(a,a; )

n even / odd

Kn0

d1

10

d2

n10

dn

An,cl.An,qu. An(1, . . . , n),

constant K set by normalisation

An,cl. = enSinst./ classical contribution

An,qu. quantum fluctuations (imported from single well): Gs.w.(0, 0; t) 1

sint

An,qu. ni

1sin(i(i+1 i))

ni

e(i+1i)/2 e/2

G(a,a; )

n even / odd

KnenSinst./e/2

n/n! 0

d1

10

d2

n10

dn

=

n even / odd

e/21

n!

KeSinst./

n

Using ex = n=0 xn/n!, N.B. non-perturbative in !G(a, a; ) Ce/2 cosh

KeSinst./

G(a,a; ) Ce/2 sinh

KeSinst./

Lecture Notes October 2005

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Lecture XIII 39

Consistency check: main contribution from

n = n

n nXn/n!

nXn/n!

= X= KeSinst./

no. per unit time, n/ exponentially small, and indep. of , i.e. dilute gas

h

S

A

SA

x

Oscillator StatesExact States

Physical interpretation: For infinite barrier two independent oscillators,coupling splits degeneracy symmetric/antisymmetric

G(a,a; ) a|SeS/S| a + a|AeA/A| a

|a|S|2

= a|SS| a =

C

2 , |a|A|2

= a|AA| a =

C

2

Setting: A/S = /2 /2

G(a,a; ) C

2

e()/2 e(+)/2

= Ce/2

cosh(/)sinh(/)

.

Remarks:

(i) Legitimacy? How do (neglected) terms O(2) compare to ?

In fact, such corrections are bigger but act equally on |S and |A

i.e. =Ke

Sinst./

is dominant contribution to splitting

V(q)

V(q)

q

q q

qm

m

(ii) Unstable States and Bounces: survival probability: G(0, 0; t)? No even/odd effect:

G(0, 0; ) = Ce/2

exp KeSinst/ =it= Ceit/2 exp 2 tDecay rate: |K|eSinst/ (i.e. K imaginary) N.B. factor of 2

Lecture Notes October 2005